Problem: What is the inverse of the function $h(x)=\dfrac{5x-3}{x-1}$ ? $ h^{-1}(x) =$
Let's start by replacing $h(x)$ with $y$. $y=\dfrac{5x-3}{x-1}$ Now let's swap $x$ and $y$ and solve for $y$. $\dfrac{5y-3}{y-1}=x$ [Why do we swap x and y?] $\begin{aligned} \dfrac{5y-3}{y-1}&=x \\\\ 5y-3&=x(y-1) \\\\ 5y-3&=xy-x \\\\ 5y-xy&=3-x \\\\ y(5-x)&=3-x \\\\ y&=\dfrac{3-x}{5-x} \end{aligned}$ In conclusion, this is the inverse function: $h^{-1}(x)=\dfrac{3-x}{5-x}$ [I saw someone solve this problem by originally solving for x. Were they wrong?]